y = x (x-3)4
Solution;
Given;Given;Given;
y=x(x−3)4y=x(x-3)^4y=x(x−3)4
ddx[(x−3)4x]\frac{d}{dx}[(x-3)^4x]dxd[(x−3)4x]
Applying the product rule;
ddx[(x−3)4].x+(x−3)4ddx[x]\frac{d}{dx}[(x-3)^4].x+(x-3)^4\frac{d}{dx}[x]dxd[(x−3)4].x+(x−3)4dxd[x]
We have;
4(x−3)3[ddx(x−3)].x+(x−3)4.14(x-3)^3[\frac{d}{dx}(x-3)].x+(x-3)^4.14(x−3)3[dxd(x−3)].x+(x−3)4.1
Differentiate the remaining part as;
4(x−3)3[1−0].x+(x−3)44(x-3)^3[1-0].x+(x-3)^44(x−3)3[1−0].x+(x−3)4
Rewrite as;
y′=(5x−3)(x−3)3y'=(5x-3)(x-3)^3y′=(5x−3)(x−3)3
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments